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## Chapter 5:

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**Chapter 5:**DISCRETE RANDOM VARIABLES AND THEIR PROBABILITY DISTRIBUTIONS**RANDOM VARIABLES**• Discrete Random Variable • Continuous Random Variable**RANDOM VARIABLES cont.**Table 5.1Frequency and Relative Frequency Distribution of the Number of Vehicles Owned by Families.**RANDOM VARIABLES cont.**• Definition • A random variable is a variable whose value is determined by the outcome of a random experiment.**Discrete Random Variable**• Definition • A random variable that assumes countable values is called a discrete random variable.**Examples of discrete random variables**• The number of cars sold at a dealership during a given month • The number of houses in a certain block • The number of fish caught on a fishing trip • The number of complaints received at the office of an airline on a given day • The number of customers who visit a bank during any given hour • The number of heads obtained in three tosses of a coin**Continuous Random Variable**• Definition • A random variable that can assume any value contained in one or more intervals is called a continuous random variable.**Examples of continuous random variables**• The height of a person • The time taken to complete an examination • The amount of milk in a gallon (note that we do not expect a gallon to contain exactly one gallon of milk but either slightly more or slightly less than a gallon.) • The weight of a fish • The price of a house**PROBABLITY DISTRIBUTION OF A DISCRETE RANDOM VARIABLE**• Definition • The probability distribution of a discrete random variable lists all the possible values that the random variable can assume and their corresponding probabilities.**Example 5-1**• Recall the frequency and relative frequency distributions of the number of vehicles owned by families given in Table 5.1. That table is reproduced below as Table 5.2. Let x be the number of vehicles owned by a randomly selected family. Write the probability distribution of x.**Table 5.2 Frequency and Relative Frequency Distributions of**the Vehicles Owned by Families**Solution 5-1**Table 5.3Probability Distribution of the Number of Vehicles Owned by Families**Two Characteristics of a Probability Distribution**• The probability distribution of a discrete random variable possesses the following two characteristics. • 0 ≤ P (x) ≤ 1 for each value of x • ΣP (x) = 1**Figure 5.1 Graphical presentation of the probability**distribution of Table 5.3. 0 1 2 3 4**Each of the following tables lists certain values of x and**their probabilities. Determine whether or not each table represents a valid probability distribution. Example 5-2**Solution 5-2**• No • Yes • No**Example 5-3**• The following table lists the probability distribution of the number of breakdowns per week for a machine based on past data.**Example 5-3**• Present this probability distribution graphically. • Find the probability that the number of breakdowns for this machine during a given week is • exactly 2 ii. 0 to 2 • more than 1 iv. at most 1**Solution 5-3**• Let x denote the number of breakdowns for this machine during a given week. Table 5.4 lists the probability distribution of x.**Table 5.4 Probability Distribution of the Number of**Breakdowns**Figure 5.2 Graphical presentation of the probability**distribution of Table 5.4. 0 1 2 3**Solution 5-3**(b) i. P (exactly 2 breakdowns) = P (x= 2) = .35 ii. P (0 to 2 breakdowns) = P (0 ≤x≤ 2) = P (x= 0) + P (x= 1) + P (x= 2) = .15 + .20 + .35 = .70 iii. P (more then 1 breakdown) = P (x> 1) = P (x= 2) + P (x= 3) = .35 +.30 = .65 iv. P (at most one breakdown) = P (x≤ 1) = P (x= 0) + P (x= 1) = .15 + .20 = .35**Example 5-4**• According to a survey, 60% of all students at a large university suffer from math anxiety. Two students are randomly selected from this university. Let x denote the number of students in this sample who suffer from math anxiety. Develop the probability distribution of x.**Solution 5-4**• Let us define the following two events:N = the student selected does not suffer from math anxiety M = the student selected suffers from math anxietyP (x = 0) = P(NN) = .16P (x = 1) = P(NM or MN) = P(NM) + P(MN) • = .24 + .24 = .48P (x = 2) = P(MM) = .36**Table 5.5 Probability Distribution of the Number of Students**with Math Anxiety in a Sample of Two Students**MEAN OF A DISCRETE RANDOM VARIABLE**The mean of a discrete variablex is the value that is expected to occur per repetition, on average, if an experiment is repeated a large number of times. It is denoted by µ and calculated as µ = Σx P (x) The mean of a discrete random variable x is also called its expected value and is denoted by E (x); that is, E (x) = Σx P (x)**Example 5-5**• Recall Example 5-3. The probability distribution Table 5.4 from that example is reproduced on the next slide. In this table, x represents the number of breakdowns for a machine during a given week, and P (x) is the probability of the corresponding value of x. • Find the mean number of breakdown per week for this machine.**Table 5.4 Probability Distribution of the Number of**Breakdowns**Table 5.6 Calculating the Mean for the Probability**Distribution of Breakdowns Solution 5-5 The mean is µ =Σx P (x) = 1.80**STANDARD DEVIATION OF A DISCRETE RANDOM VARIABLE**• The standard deviation of a discrete random variablex measures the spread of its probability distribution and is computed as**Example 5-6**• Baier’s Electronics manufactures computer parts that are supplied to many computer companies. Despite the fact that two quality control inspectors at Baier’s Electronics check every part for defects before it is shipped to another company, a few defective parts do pass through these inspections undetected. Let x denote the number of defective computer parts in a shipment of 400. The following table gives the probability distribution of x.**Example 5-6**Compute the standard deviation of x.**Solution 5-6**Table 5.7Computations to Find the Standard Deviation**Example 5-7**• Loraine Corporation is planning to market a new makeup product. According to the analysis made by the financial department of the company, it will earn an annual profit of $4.5 million if this product has high sales and an annual profit of $ 1.2 million if the sales are mediocre, and it will lose $2.3 million a year if the sales are low. The probabilities of these three scenarios are .32, .51 and .17 respectively.**Example 5-7**• Let x be the profits (in millions of dollars) earned per annum by the company from this product. Write the probability distribution of x. • Calculate the mean and the standard deviation of x.**Solution 5-7**• The following table lists the probability distribution of x.**Table 5.8 Computations to Find the Mean and Standard**Deviation**FACTORIALS AND COMBINATIONS**• Factorials • Combinations • Using the Table of Combinations**Factorials**• Definition • The symbol n!, read as “n factorial”, represents the product of all the integers from n to 1. in other words, • n! = n(n - 1)(n – 2)(n – 3). . . 3 . 2 . 1 • By definition, • 0! = 1**Example**Evaluate the following: • 7! • 10! • (12 – 4)! • (5 – 5)!**Solution**• 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 • 10! = 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 3,628,800 • (12 – 4)! = 8! = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1 = 40,320 d) (5 – 5)! = 0! = 1**Combinations**• Definition • Combinations give the number of ways x elements can be selected from n elements. The notation used to denote the total number of combinations is • which is read as “the number of combinations of n elements selected x at a time.”**Combinations cont.**n denotes the total number of elements = the number of combinations of n elements selected x at a time x denotes the number of elements selected per selection**Combinations cont.**• Number of Combinations • The number of combinations for selecting x from n distinct elements is given by the formula**Example 5-13**• An ice cream parlor has six flavors of ice cream. Kristen wants to buy two flavors of ice cream. If she randomly selects two flavors out of six, how many combinations are there?